When solving an absolute value, for example 2|x+3| = 24 is it okay to distribute the 2 (2x+6 = 24) and then set up your two cases or is the only correct way to divide by 2 (x+3=12) and then set up your cases? I use the distribution method in my class but my teacher told us to use the division method so I asked her if distribution was okay and she said she didn't know.

Sep 10th 2013, 06:58 PM

chiro

Re: Question when solving absolute values

Hey alext180.

You can divide both sides b y 2 in this case but you can not in general distribute what is in the absolute value expression. So you can simply to |x+3| = 12 but then you need to evaluate when the expression inside |.| is < 0 and >= 0 separately.

Sep 11th 2013, 09:06 AM

johng

Re: Question when solving absolute values

Hi,
Yes, you can do what you suggest. Use the fact that |ab| = |a||b|. (If you don't know this, you can prove it by considering all possibilities for the signs of a and b.) So 2|x+3|=|2||x+3|=|2(x+3)|=|2x+6|=24, etc. Notice if the problem were -2|x+3|=24, you could still "distribute" if you're careful. -2|x+3|=-|2||x+3|=-|2x+6|=24, an equation with no solution.